Question

The maximum value of $f\left(x\right)=\frac{3x^2+9x+17}{3x^2+9x+7}$ is $5k+1,$ then the value of $k$ is ........

Answer 1

Let $y=\frac{3x^2+9x+17}{3x^2+9x+7}=1+\frac{10}{3x^2+9x-7}$
Now, $3x^2+9x+7=3(x^2+3x)+7$
$=3{(x+\frac{3}{2})}^2+\frac{1}{4}\ge \frac{1}{4}$ for all $x\in R$.
Maximum value of $\frac{10}{3x^2+9x-7}$ is 40.
Maximum value of y is $1+40=41$
$\therefore 5k+1=41$
$\Rightarrow k=8$